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(20 pts) When 3.018 moles of chlorine reacts with excess aluminum, how many moles of
aluminum chloride are formed?
2 Al + 3 Cl2 → 2 AICI3:

Answer :

znk

Answer:

[tex]\large \boxed{\text{2.012 mol AlCl}_{3}}[/tex]

Explanation:

             2Al + 3Cl₂ → 2AlCl₃

n/mol:            3.018

The molar ratio is 2 mol AlCl₃:3 mol Cl₂.

[tex]\text{Moles of AlCl}_{3} = \text{ 3.018 mol Cl}_{2} \times \dfrac{\text{2 mol AlCl}_{3}}{\textbf{3 mol Cl}_{2}}\\\\= \text{2.012 mol AlCl}_{3}\\\text{The reaction produces $\large \boxed{\textbf{2.012 mol AlCl}_{3}}$}[/tex]

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